It is a relatively simple exercise to prove that a well-ordered set is order-isomorphic to a subset of $\mathbb R$ (under the usual ordering) if and only if it is countable. You can say that $\mathbb R$ is "too small" to contain any uncountable well-ordered sets.

So my question is, can you embed bigger well-ordered sets in the long line? For those who don't know, the long line can be constructed by taking the minimal uncountable well-ordered set (i.e. $\omega_1$) and taking its Cartesian product with $[0,1)$ under the dictionary order. So obviously $\omega_1$ itself is emebeddable in the long line, just by taking the left endpoints of all the intervals $[0,1)$. But can you embed bigger uncountable ordinals, and if so how big? I'm guessing that you may be able to embed all well-ordered sets with cardinality less than or equal to $\aleph_1$, the cardinality of the set of countable ordinals.

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    $\begingroup$ You should look into surreal numbers. They have a similar relationship to ordinals as the reals do to the natural numbers (except the surreals are not topologically complete, nor do they have any topologically complete field extension). $\endgroup$ – Paul Sinclair Jan 9 '18 at 17:11

Not that many, really.

The thing about the "long line" is that every proper initial segment is "just a line" (read: an interval of $\Bbb R$). So every proper initial segment can only house countable ordinals.

So no uncountable ordinal greater than $\omega_1$ can be embedded in the long line; and since the long line trivially embeds $\omega_1$, by picking the end-points of the intervals, we get that the only uncountable ordinal that can be embedded in the long line is $\omega_1$ itself.

  • $\begingroup$ Thanks for your answer. What about bigger versions of the long line, i.e. versions that are constructed using a Cartesian product with a bigger uncountable ordinal? Do you ever get more ordinals than you put in, i.e. can you embed larger ordinals than the ordinal you use for the Cartesian product? $\endgroup$ – Keshav Srinivasan Jan 9 '18 at 14:25
  • $\begingroup$ I don't have time or inclination to verify the details. Off the bat, I would say that the supremum of the ordinals that embed into $[0,1)\times\alpha$ is $\alpha\cdot\omega_1$, and that supremum is not necessarily a maximum. (Note that $\alpha\cdot\beta$ is the order type of $\beta\times\alpha$.) $\endgroup$ – Asaf Karagila Jan 9 '18 at 14:29

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